Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 126507 by joki last updated on 21/Dec/20

$$\int \frac{1}{(\mathrm{x}-2)\sqrt{{\mathrm{x}}^2-4\mathrm{x}+1)}}\mathrm{dx}$$

Answered by Dwaipayan Shikari last updated on 21/Dec/20

$$\int \frac{1}{(x-2)\sqrt{{(x-2)}^2-{\left(\sqrt{3}\right)}^2}}dx$$$$={sec}^{-1}\frac{x-2}{\sqrt{3}}+C$$

Commented by joki last updated on 21/Dec/20

$$\mathrm{can}\mathrm{you}\mathrm{detail}\mathrm{your}\mathrm{answer}\mathrm{sir}.\mathrm{thanks}$$

Answered by Ar Brandon last updated on 21/Dec/20

$$\mathcal{I}=\int \frac{\mathrm{dx}}{(\mathrm{x}-2)\sqrt{{\mathrm{x}}^2-4\mathrm{x}+8}}$$$$\mathrm{Let}\frac{1}{\mathrm{x}-2}=\mathrm{u}\Rightarrow -\frac{\mathrm{dx}}{{(\mathrm{x}-2)}^2}=\mathrm{du}\Rightarrow \mathrm{dx}=-\frac{\mathrm{du}}{{\mathrm{u}}^2}$$$${\mathrm{x}}^2-4\mathrm{x}+8={(\mathrm{x}-2)}^2+4$$$$\mathcal{I}=-\int \frac{\mathrm{u}}{\sqrt{\frac{1}{{\mathrm{u}}^2}+4}}\cdot \frac{\mathrm{du}}{{\mathrm{u}}^2}=\mp \int \frac{\mathrm{du}}{\sqrt{1+{4\mathrm{u}}^2}}$$$$=\mp \ln \mid 2\mathrm{u}+\sqrt{1+{4\mathrm{u}}^2}\mid +\mathrm{C}=\mp \ln \mid \frac{2+\sqrt{{\mathrm{x}}^2-4\mathrm{x}+8}}{\mathrm{x}-2}\mid +\mathrm{C}$$

Answered by Ar Brandon last updated on 21/Dec/20

$$\mathcal{I}=\int \frac{\mathrm{dx}}{(\mathrm{x}-2)\sqrt{{\mathrm{x}}^2-4\mathrm{x}+1}}$$$$\mathrm{Let}\frac{1}{\mathrm{x}-2}=\mathrm{u}\Rightarrow -\frac{\mathrm{dx}}{{(\mathrm{x}-2)}^2}=\mathrm{du}\Rightarrow \mathrm{dx}=-\frac{\mathrm{du}}{{\mathrm{u}}^2}$$$${\mathrm{x}}^2-4\mathrm{x}+1={(\mathrm{x}-2)}^2-3$$$$\mathcal{I}=-\int \frac{\mathrm{u}}{\sqrt{\frac{1}{{\mathrm{u}}^2}-3}}\cdot \frac{\mathrm{du}}{{\mathrm{u}}^2}=\mp \int \frac{\mathrm{du}}{\sqrt{1-{3\mathrm{u}}^2}}$$$$=\mp \frac{1}{\sqrt{3}}\mathrm{Arcsin}\left(\sqrt{3}\mathrm{u}\right)+\mathrm{C}=\mp \frac{1}{\sqrt{3}}\mathrm{Arcsin}\left(\frac{\sqrt{3}}{\mathrm{x}-2}\right)+\mathrm{C}$$

Answered by MJS_new last updated on 21/Dec/20

$$\int \frac{dx}{(x-2)\sqrt{x^2-4x+1}}=$$$$[t=\frac{x-2+\sqrt{x^2-4x+1}}{\sqrt{3}}\to dx=\frac{\sqrt{3(x^2-4x+1}}{x-2+\sqrt{x^2-4x+1}}]$$$$=\frac{2}{\sqrt{3}}\int \frac{dt}{t^2+1}=\frac{2}{\sqrt{3}}\arctan t=$$$$=\frac{2}{\sqrt{3}}\arctan \frac{x-2+\sqrt{x^2-4x+1}}{\sqrt{3}}+C$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com